Question: A certain circle can be represented by the following equation. $x^2+y^2-6x-6y+2=0$ What is the center of this circle ? $($
Solution: The strategy We can find the center and radius of a circle by rewriting the given equation in the form of the standard equation of a circle. [What is the standard equation of the circle?] In order to do this, we take the following steps. Complete the square for both the $x^2$ and $y^2$ terms. [How do we complete the square?] Write the equation in the standard form of the circle. Completing the squares $\begin{aligned}x^2+y^2-6x-6y+2&=0\\\\ x^2+y^2-6x-6y&=-2\\\\ (x^2-6x)+(y^2-6y)&=-2 \text{(rearrange terms)}\\\\ (x^2-6x{+9})+(y^2-6y{+9})&=-2{+9}{+9}\end{aligned}$ Notice that we must add ${9}$ and ${9}$ on the right side of the equation, since we added them to the left side of the equation. [How did we get 9 and 9?] Writing the equation in standard form $\begin{aligned}(x^2-6x{+9})+(y^2-6y{+9})&=-2{+9}{+9}\\\\ (x-3)^2+(y-3)^2&=16\\\\ (x-3)^2+(y-3)^2&=4^2\end{aligned}$ Since the equation is now in the standard form, we can conclude that this circle is centered at $(3,3)$ and has a radius of $4$ units. Summary The circle is centered at $(3,3)$. The circle has a radius of $4$ units.